Solution

A body is travelling along a straight line with constant acceleration \(\quantity{a}{m\,s^{-2}}\). The body passes a fixed point \(O\) on the line with velocity \(\quantity{u}{m\,s^{-1}}\). Between \(\quantity{4}{s}\) and \(\quantity{5}{s}\) after passing \(O\) the body travels \(\quantity{10}{m}\). Between \(\quantity{6}{s}\) and \(\quantity{7}{s}\) after passing \(O\) the body travels \(\quantity{12}{m}\).

Calculate the values of \(u\) and \(a\).

We are given distances at various times so the most useful equation of motion is going to be \[s=ut+\frac{1}{2}at^2.\]

We can use the equation to find the displacements at \(\quantity{5}{s}\) and \(\quantity{4}{s}\). \[s_5=5u+\frac{25}{2}a\]\[s_4=4u+\frac{16}{2}a\] Subtracting gives us the distance travelled in this period which we are told is \(10\). We can do the same for the period \(\quantity{6}{s}\) to \(\quantity{7}{s}\). \[10=u+\frac{25-16}{2}a=u+4.5a\]\[12=u+\frac{49-36}{2}a=u+6.5a\] We now can subtract these equations and solve for \(a\), \[-2=-2a\]\[a=1\] and substitute this value into one of the above equations to find \(u\). \[10=u+4.5\]\[u=5.5\]

Alternatively, we could use a graphical approach to visually achieve the same solution. If we draw a velocity-time graph, the area underneath is the distance travelled.

From the information we are given, the difference between the two shaded areas is \(\quantity{2}{m}\). In the period of time between \(\quantity{4}{s}\) and \(\quantity{6}{s}\) the velocity increases by \(2a\). So the difference between the shaded areas is a rectangle of base \(1\) and height \(2a\). Since we know this difference is \(2\) we have \[2a\times1=2\]\[a=1\]